Question: Simplify and expand the following expression: $ \dfrac{a - 10}{2a - 10}-\dfrac{4a + 5}{5a + 1} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(2a - 10)(5a + 1)$ Multiply the first term by $\dfrac{5a + 1}{5a + 1}$ $ \begin{align*} \dfrac{a - 10}{2a - 10} \times \dfrac{5a + 1}{5a + 1} & = \dfrac{(a - 10)(5a + 1)}{(2a - 10)(5a + 1)} \\ & = \dfrac{5a^2 - 49a - 10}{(2a - 10)(5a + 1)}\end{align*} $ Multiply the second term by $\dfrac{2a - 10}{2a - 10}$ $ \begin{align*} \dfrac{4a + 5}{5a + 1} \times \dfrac{2a - 10}{2a - 10} & = \dfrac{(4a + 5)(2a - 10)}{(5a + 1)(2a - 10)} \\ & = \dfrac{8a^2 - 30a - 50}{(5a + 1)(2a - 10)}\end{align*} $ Now we have: $ = \dfrac{5a^2 - 49a - 10}{(2a - 10)(5a + 1)} - \dfrac{8a^2 - 30a - 50}{(5a + 1)(2a - 10)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{5a^2 - 49a - 10 - (8a^2 - 30a - 50)}{(2a - 10)(5a + 1)} $ $ = \dfrac{5a^2 - 49a - 10 - 8a^2 + 30a + 50}{(2a - 10)(5a + 1)} $ $ = \dfrac{-3a^2 - 19a + 40}{(2a - 10)(5a + 1)}$ Expand the denominator: $ = \dfrac{-3a^2 - 19a + 40}{10a^2 - 48a - 10}$